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Chapter 5

Chapter 5. Gases. P, n, T, R. g. V. n. Calculate the Volume Occupied by 637 g of SO 2 (MM 64.07) at 6.08 x 10 3 mmHg and –23 °C . Given: Find:. m SO2 = 637 g, P = 6.08 x 10 3 mmHg, t = −23 °C, V , L. Solution Map: Relationships:. 1 atm = 760 mmHg

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Chapter 5

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  1. Chapter 5 Gases

  2. P, n, T, R g V n Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C. Given: Find: mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C, V, L Solution Map: Relationships: 1 atm = 760 mmHg T(K) = t(°C) + 273, 1 mol SO2 = 64.07 g Solution:

  3. V, m P, n, T, R d V 1 atm = 760 mmHg, T(K) = t(°C) + 273 Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g Given: Find: m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300. K density,g/L m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C, density,g/L Solution Map: Relationships: Solution:

  4. Molar Mass of a Gas • One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

  5. n, m P, V, T, R MM n 1 atm = 760 mmHg, T(K) = t(°C) + 273 Calculate the Molar Mass of a Gas with Mass 0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg. Given: Find: m = 0.311g, V = 0.225 L, P = 1.1658 atm, T = 328 K Molar Mass,g/mol m=0.311g, V=0.225 L, P=886 mmHg, t=55°C, Molar Mass,g/mol Solution Map: Relationships: Solution:

  6. Write a solution map: Information: Given: V = 0.225 L, P = 886 mmHg, t = 55 °C, m = 0.311 g Find: molar mass, (g/mol) Equation: PV = nRT; MM = mass/moles mass Example:A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass. P,V,T,R n Molar Mass PV = nRT When using the ideal gas equation, the units of V must be L; and the units of Pmust be atm, or you will have to convert. The units of T must be kelvin, K.

  7. What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C? n, m P, V, T, R MM n 1 atm = 760 torr, T(K) = t(°C) + 273 Given: Find: m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K, molar mass,g/mol m=12.0 g, V= 197 L, P=380 torr, t=127°C, molar mass,g/mol Solution Map: Relationships: Solution:

  8. Partial Pressure • Each gas in the mixture exerts a pressure independent of the other gases in the mixture. • The pressure of a component gas in a mixture is called a partial pressure. • The sum of the partial pressures of all the gases in a mixture equals the total pressure. • Dalton’s law of partial pressures. • Ptotal = Pgas A + Pgas B + Pgas C +...

  9. A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture. Ptot, PHe, PNe PAr Given: Find: PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg PAr, mmHg Solution Map: Relationships: PAr = Ptot – (PHe + PNe) Ptot= Pa + Pb + etc. Solution:

  10. Finding Partial Pressure • To find the partial pressure of a gas, multiply the total pressure of the mixture by the fractional composition of the gas. • For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be: PHe = (0.800)(1.0 atm) = 0.80 atm • Fractional composition = percentage divided by 100.

  11. The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be Calculated Using the Ideal Gas Law 11

  12. Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe. nXe, V, T, R PXe Ptot, PXe PNe Given: Find: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm Solution Map: Relationships: Solution:

  13. Deep Sea Divers and Partial Pressure • It is also possible to have too much O2, a condition called oxygen toxicity. • PO2 > 1.4 atm. • Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions. • It is also possible to have too much N2, a condition called nitrogen narcosis. • Also known as rapture of the deep. • When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases. • At a depth of 55 m, the partial pressure of O2 is 1.4 atm. • Divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air.

  14. Partial Pressure vs. Total Pressure At a depth of 30 m, the total pressure of air in the divers lungs, and the partial pressure of all the gases in the air, are quadrupled!

  15. Gas Stoichiometry

  16. Write a solution map: Information: Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 305 K Find: VO2, L Equation: PV = nRT Conversion Factors: 1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2 P,T,R Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K. g KClO3 nKClO3 nO2 VO2 PV=nRT When using the ideal gas equation, the units of V must be L; and the units of Pmust be atm, or you will have to convert. The units of T must be kelvin, K.

  17. Apply the solution map: Find moles of O2 made. Information: Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 305 K Find: VO2, L Equation: PV = nRT Conversion Factors: 1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2 Solution Map: g → mol KClO3→ mol O2 → L O2 Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.

  18. Apply the solution map: Convert the units. Information: Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 305 K, nO2 = 3.60 moles Find: VO2, L Equation: PV = nRT Conversion Factors: 1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2 Solution Map: g → mol KClO3→ mol O2 → L O2 Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.

  19. Apply the solution map: Information: Given: 294 g KClO3 PO2 = 0.99342 mmHg, TO2 = 305 K, nO2 = 3.60 moles Find: VO2, L Equation: PV = nRT Conversion Factors: 1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2 Solution Map: g → mol KClO3→ mol O2 → L O2 Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.

  20. g HgO mol HgO mol O2 1 atm = 760 mmHg, HgO = 216.59 g/mol 2 mol HgO : 1 mol O2 P, n, T, R V Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g), Continued Given: Find: mHgO = 10.0g, P=0.750 atm, T=313 K VO2,L nO2 = 0.023085 mol, P = 0.750 atm, T = 313 K VO2,L Solution Map: Relationships: Solution:

  21. L∙atm mol∙K (1.00 atm) x V = (1.00 moles)(0.0821 )(273 K) Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP. P x V = n x R x T • 1 mole of any gas at STP will occupy 22.4 L. • This volume is called the molar volume and can be used as a conversion factor. • As long as you work at STP. 1 mol  22.4 L V = 22.4 L

  22. Molar Volume There is so much empty space between molecules in the gas state that the volume of the gas is not effected by the size of the molecules (under ideal conditions).

  23. L H2 mol H2 mol H2O g H2O Example —How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP?O2(g) + 2 H2(g) → 2 H2O(g) Given: Find: VH2 = 1.24 L, P = 1.00 atm, T = 273 K massH2O,g Solution Map: Relationships: H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP 2 mol H2O : 2 mol H2 Solution: 23

  24. Write a solution map: Information: Given: 1.24 L H2 Find: g H2O Conversion Factors: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? L H2 mol H2 mol H2O g H2O

  25. Apply the solution map: Information: Given: 1.24 L H2 Find: g H2O Conversion Factors: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L Solution Map: L → mol H2→ mol H2O → g H2O Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?

  26. g HgO mol HgO mol O2 L O2 Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g), Continued Given: Find: mHgO = 10.0 g, P = 1.00 atm, T = 273 K VO2,L Solution Map: Relationships: HgO = 216.59 g/mol, 1 mol = 22.4 L at STP 2 mol HgO : 1 mol O2 Solution: 26

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